Unequal mass spring system

Im revising for my end of year exams and this was a question i got wrong in a test we had this year and unforunately there arent worked answers. I sorta get close but not quite there.

The answer is A, which i get, sorta
[tex] F = ma = 2kx [/tex]
As per newtons third law (the hint) saying that the pulling and pushing force in turn create a pushing and pulling force, thus doubling the force. So using
[tex] s = ut +1/2 at^2[/tex]
Now the different masses will go different distances so if we take a ratio of masses, ie the smaller one going further we get
[tex]\frac{4x}{5} = \frac{kxt^2}{m}[/tex]
[tex] \frac{4}{5} = \frac{kt^2}{m}[/tex]

here im trying to use t as period and thus get f via T = 1/f
Now this works fine, you rearange for t and you get the answer A. However one thing that puzzels me is: if this is supposed to be the period, surely we should consider one whole oscilation, that is to say the distance would infact be 8/5, which throws a spanner in the works. Any ideas?

Staff: Mentor

well the centre of mass, and i realise now im missing something obvious, but how does that help me?

CM = 4/5L or x+L depending on length (from the small mass) right? and 1/5 away from the big one? please help im willing to look stupid and learn if i can get to the bottom of this, itsb een bugging me all day :P

well the centre of mass, and i realise now im missing something obvious, but how does that help me?

CM = 4/5L or x+L depending on length (from the small mass) right? and 1/5 away from the big one? please help im willing to look stupid and learn if i can get to the bottom of this, itsb een bugging me all day :P

I assume you know how the force on each mass depends on the Δx of the spring. Imagine adding a rigid clamp to the spring at that point that does not move. That would have no effect on the motion on either side of the clamp. You could cut one side of the spring off, and the other side with its mass would continue to oscillate with the same frequency it had before the clamp was added. What does cutting a spring do to its behavior?

Yeh but thats what ive done isnt it? im asking if my stuff is right, obviously your posting because its not and i would like to know which part. whilst i appriciate the riddle aspect and i dont get something for nothing, ive presented working here. I mean i get the right answer, but theres the discrepincy about the distance travelled still.

Staff: Mentor

Yes, you did show some work, but I don't understand it. Something about Newton's III implying a doubling of the force? Better to start over with the hints we gave.

The center of mass does not move. So pretend two smaller springs are fixed at that point and each has a mass at the end. Pick one of those springs. What's its spring constant? What's its mass? What's its frequency? It's just a mass on the end of a spring. (I assume you know how to find the frequency of oscillation of a single mass on a spring.)

Well no offence but my work isnt that hard to follow. I understand your point . Youre saying the force is simply kx where x is a different distance for each mass, as i worked out before. I tried this way but it leads to the wrong answer.

I used s = ut + 1/2at^2 with ut = 0 as initial velocity is zero. Then if F = ma = kx then 1/5 (for large mass) = (1/2)(k/4m)t^2 which gives root 8/5 not root 4/5. Furthermore the point which i raised as my problem but which hasnt been addressed is should s be x or 2x? Ie the distance traveled in one period.

Whilst your hints are helpful id really like some serious numerical help here as ive done (initially anyway) what you guys suggested but im getting nowhere.

Staff: Mentor

FunkyDwarf said:

Well no offence but my work isnt that hard to follow. I understand your point . Youre saying the force is simply kx where x is a different distance for each mass, as i worked out before. I tried this way but it leads to the wrong answer.

The spring exerts the same force on each mass, that's for sure. And each mass does move a different distance as it oscillates. But the center of mass of the system is fixed.

I'm saying that an easy way to solve this problem is to treat each mass as being on its own, separate (shorter) spring, with a different spring constant than the original. Did you try that?

I used s = ut + 1/2at^2 with ut = 0 as initial velocity is zero. Then if F = ma = kx then 1/5 (for large mass) = (1/2)(k/4m)t^2 which gives root 8/5 not root 4/5. Furthermore the point which i raised as my problem but which hasnt been addressed is should s be x or 2x? Ie the distance traveled in one period.

Why are you using a kinematics formula for constant acceleration? It's hard to offer a comment when I don't see what you are trying to do.

Whilst your hints are helpful id really like some serious numerical help here as ive done (initially anyway) what you guys suggested but im getting nowhere.

Why don't you try following our suggestions? (Once you see what we're saying, you'll be done in 60 seconds.) Treat the spring as being in two pieces, connected at the center of mass. Answer these questions:
(1) How long is each spring?
(2) What's the spring constant of each spring?
(3) What's the frequency of oscillation of each mass on its spring?

The reason im using that forumula is because it contains constants i have and the variable i need. I know the distance it moves and the accleration it feels and the time variable is there. Also its not constant acceleration, the force and therefore accel changes with distance.

1.) Well its all in variables so all i can say is the distance from the equilibrium position each one moves, which is all that really matters as if you look in the possible answers theres no L, plus its logical.
2.) this doesnt come into it given the question so we just use k
3.) Again, using that formula you can get t for one period which is T and then 1/T is f.

The reason im using that forumula is because it contains constants i have and the variable i need. I know the distance it moves and the accleration it feels and the time variable is there. Also its not constant acceleration, the force and therefore accel changes with distance.

1.) Well its all in variables so all i can say is the distance from the equilibrium position each one moves, which is all that really matters as if you look in the possible answers theres no L, plus its logical.
2.) this doesnt come into it given the question so we just use k
3.) Again, using that formula you can get t for one period which is T and then 1/T is f.

What is wrong with this? I dont get what youre saying

When I first replied, you had already said you recognized that you were missing something. I will be more specific about what you are missing, and in doing so I will repeat some of what Doc Al has been saying. I will add some coments to your original post to put those comments into context.

FunkyDwarf said:

Hey guys

Im revising for my end of year exams and this was a question i got wrong in a test we had this year and unforunately there arent worked answers. I sorta get close but not quite there.

The answer is A, which i get, sorta
[tex] F = ma = 2kx [/tex]
As per newtons third law (the hint) saying that the pulling and pushing force in turn create a pushing and pulling force, thus doubling the force.

This equation is not correct. The observation is not correct. The hint about Newton's 3rd law is not about pushing vs pulling. There is no doubling of the force. By Newton's 3rd law there are two equal and opposite forces, an action-reaction pair, at each end of the spring. Each mass exerts a force on the spring and each mass experiences an equal and opposite force from the spring.

It takes forces applied in opposite directions at opposite ends of the spring to either stretch or compress the spring. When those two forces acting on the spring (not an action reaction pair- they are acting on the same object) are outward it stretches, and when they are acting inward it compresses. For simplicity of notation and because you used just x I will use x (instead of the Δx in the problem statement) to represent the elongation of the sring. The amount by which the length of the spring changes is related to the magnitude of this force by

F = kx

When x is positive, the spring is stretched and when x is negative the spring is compressed. By Newton's 3rd law, a force of this magnitude is experienced by each mass. There is no F = 2kx

So using
[tex] s = ut +1/2 at^2[/tex]

This equation is not applicable to any situation where the force and acceleration are not constant, which is the case in this problem.

Now the different masses will go different distances

This is true. They will go different distances, and it appears you have figured out the correct ratio of the distances they move.

so if we take a ratio of masses, ie the smaller one going further we get
[tex]\frac{4x}{5} = \frac{kxt^2}{m}[/tex]
[tex] \frac{4}{5} = \frac{kt^2}{m}[/tex]
here im trying to use t as period and thus get f via T = 1/f

I see no connection between this and anything you have written previously. The justification for introducing a factor of 4/5 or 5/4 into the problem is not established by writing the first of these equations, even if it does happen to reduce the "right answer." There is no relationship between the period of an oscillator and the amount by which the spring is stretched or compressed at any time during its motion.

Now this works fine, you rearange for t and you get the answer A. However one thing that puzzels me is: if this is supposed to be the period, surely we should consider one whole oscilation, that is to say the distance would infact be 8/5, which throws a spanner in the works. Any ideas?

I have no idea what you mean by the distance being 8/5, but your comment shows that you do not really know why the 5/4 factor enters into the answer to the problem.

Cheers
-G

The thing we have been trying to lead you to is recognizing that each mass experiences an equal and opposite force of magnitude

F = kx

For each mass this force tends to restore the mass to its equilibrium position. However, the x in this equation is not the displacement of the mass. To apply the formulas for a harmonic oscillator, you need to know the relationship between the force applied to a moving mass and the distance that mass moves. You recognize that one mass moves more than the other, and it is obvious from what you have done that you can write the distance each mass moves in terms of x. Your next step should be to write two equations, one for each mass, that expresses the force acting on that mass in terms of the distance that mass moves. Once you have those equations, you will have an effective spring constant (not k, but proportional to k) for each mass and then you can apply the usual equations for harmonic motion. Either of the two equations will lead you to the same conclusion about the frequency of oscillation shared by the two masses.

Thats better :) As for the last blue bit i do know why its just i didnt express it very well.

As for the rest of it im a bit stuck. We never actually did this sort of thing in classes, its a general mechanics course (well not course but sub unit) and we just have the general equations of motion to work with.

That being said i can see how you can derive the situation for just one m where f = 2pi sqrt(k/m) thats fine i get the derivation of that. But when i try an incorporate the variable distances traveled by each mass i get the displacement canceling out which is obviously useless (and probably wrong)

if (ignoring L) [tex] x = Acos(wt) [/tex] and say its the big mass so its
[tex] x = (1/5)Cos(wt) [/tex] then [tex] a = -w^2Cos(wt) = -w^2 x [/tex] as per the rules and definition of SHM (this second x of course being representative of the cos related function)However, any way i try it i still just get the basic form, instead of one involving 5/4.

Thats better :) As for the last blue bit i do know why its just i didnt express it very well.

As for the rest of it im a bit stuck. We never actually did this sort of thing in classes, its a general mechanics course (well not course but sub unit) and we just have the general equations of motion to work with.

That being said i can see how you can derive the situation for just one m where f = 2pi sqrt(k/m) thats fine i get the derivation of that. But when i try an incorporate the variable distances traveled by each mass i get the displacement canceling out which is obviously useless (and probably wrong)

if (ignoring L) [tex] x = Acos(wt) [/tex] and say its the big mass so its
[tex] x = (1/5)Cos(wt) [/tex] then [tex] a = -w^2Cos(wt) = -w^2 x [/tex] as per the rules and definition of SHM (this second x of course being representative of the cos related function)However, any way i try it i still just get the basic form, instead of one involving 5/4.

Try using new variables to represent the displacements of the two masses from their equilibrium positions, say y for the big mass and z for the small mass with directions chosen so that both are positive when the spring is stretched (y + z = x). Then write the two equations I suggested you write in the previous post.

F = F(y) = ???
F = F(z) = ???

Assume a sinusoidal solution for y and z as you do for any oscillator and figure out what ω has to be to satisfy each equation.